Quadratic Equations

To solve the quadratic equation ax^{2}+bx+c=0, we factorise the quadratic expression ax^{2}+bx+c into two factors. The solutions are then obtained by setting each of the factors equal to zero. We shall obtain two solutions.

Let’s look at some examples.

Example

Solve the equations:

i) (2x-1)(3x+2)=0

ii) 2x^{2}+x-3=0

i) (2x-1)(3x+2)=0

Either 2x-1=0

x = \frac{1}{2}

Or, 3x+2=0

x = \frac{-2}{3}

ii) 2x^{2}+x-3=0

2x^{2}+3x-2x-3=0

x(2x+3)-1(2x+3)=0

(x-1)(2x+3)=0

So, either x-1=0

x=1

Or, 2x+3=0

x = \frac{-3}{2}


Example

Solve the quadratic equation:

x^{2}-16=0

x^{2}-16=0

x^{2}=16

x = \sqrt{16}

x=\pm 4

Or, we can do this by factors

x^{2}-16=0

(x+4)(x-4)=0

Either x+4=0

x=-4

Or, x-4=0

x=4


Example

Solve the equation 6x+7 = \frac{5}{x}

6x+7 = \frac{5}{x}

We multiply both sides by x.

x(6x+7)=5

Then, we expand the bracket and simplify.

6x^{2}+7x=5

6x^{2}+7x-5=0

Factorise: 6x^{2}+10x-3x-5=0

2x(3x+5)-1(3x+5)=0

(2x-1)(3x+5)=0

Solve: x = \frac{1}{2} or x = \frac{-5}{3}


The quadratic expression ax^{2}+bx+c is not always factorisable in rational numbers. This situation arises when the quantity b^{2}-4ac, called the discriminant, is not a perfect square, meaning it has no rational square roots.

We then solve the quadratic equation ax^{2}+bx+c=0, using the formula x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}

Consider the quadratic equation:

2x^{2}-7x+1=0

For the equation:

a = 2, b = -7 and c = 1.

We substitute these numbers into the formula above. So:

x=\dfrac{-\left( -7\right) \pm \sqrt{\left( -7\right) ^{2}-4\left( 2\right) \left( 1\right) }}{2\left( 2\right) }

=\dfrac{7\pm \sqrt{49-8}}{4}

We obtain x = \frac{7- \sqrt{41}}{4}

= 0.49 (3 significant figures)

Or, x = \frac{7 + \sqrt{41}}{4}

= 3.35 (3 significant figures)

Let’s look at some examples.

Example

Solve the quadratic equation:

5x^{2}-14x+3=0

Giving answers correct to three significant figures

5x^{2}-14x+3=0

a=5, b=-14, c=3

x=\dfrac{14\pm \sqrt{14^{2}-4\left( 5\right) \left( 3\right) }}{2\left( 5\right) }

=\dfrac{14\pm \sqrt{136}}{10}

x=0.234 or x=2.57, each correct to three significant figures. 


Example

Find the possible values of x for which:

2x^{2}-7x+3=0

2x^{2}-7x+3=0

Here, a=2, b=-7 and c=3.

We substitute these numbers into x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}

x=\dfrac{-\left( -7\right) \pm \sqrt{\left( -7\right) ^{2}-4\left( 2\right) \left( 3\right) }}{2\left( 2\right) }

=\dfrac{7\pm \sqrt{49-24}}{4}

=\dfrac{7\pm \sqrt{25}}{4}

=\dfrac{7\pm 5}{4}

x = \frac{7-5}{4} = \frac{1}{2}

Or, x = \frac{7+5}{4} = 3

The possible values of x are \frac{1}{2} and 3.

Note: In fact, the quadratic equation 2x^{2}-7x+3=0 can be solved using factorisation.

2x^{2}-7x+3=0

2x^{2}-6x-x+3=0

2x(x-3)-1(x-3)=0

(2x-1)(x-3)=0

Either 2x-1=0, giving x = \frac{1}{2}

Or, x-3=0, giving x=3.


Example

A solid cuboid of length x metres has a square cross-section of side (x-3) metres. Write down, in terms of x, expressions for:

i) The surface are of the square cross-section.

ii) The surface area of one of the longer faces of the cuboid.

Given that the total surface area of the cuboid is 39 square metres, write down an equation in x and show that it reduces to:

2x^{2}-8x-7=0

Find the two values of x which satisfies this equation, giving your answers correct to three significant figures. Hence right down the dimensions of the cuboid.

i) Surface are of the cross-section = (x-3)(x-3)

= (x^{2}-6x+9)\:m^{2}

ii) Surface area of one of the longer faces

= x(x-3)

= (x^{2}-3x)\:m^{2}

Total surface area of the cuboid:

= 2(x^{2}-6x+9)+4(x^{2}-3x)

= 2x^{2}-12x+18+4x^{2}-12x

= 6x^{2}-24x+18

Given the total surface area = 39\:m^{2}:

6x^{2}-24x+18=39

6x^{2}-24x+18-39=0

6x^{2}-24x-21=0

Dividing by the common factor 3, we obtain:

2x^{2}-8x-7=0, as required.

We use the formula to find x

x=\dfrac{8\pm \sqrt{64-4\left( 2\right) \left( -7\right) }}{2\left( 2\right) }

x=\dfrac{8\pm \sqrt{120}}{4}

x=-0.739 or x=4.74, each to three significant figures.

The dimensions of the cuboid are 4.74 m, 1.74 m


Example

Solve the equation:

\frac{x-5}{x-2}+1=\frac{3x}{x-1}

\frac{x-5}{x-2}+1=\frac{3x}{x-1}

We can simplify the left hand side, then cross-multiply.

\frac{x-5}{x-2}+1=\frac{x-5+1(x-2)}{x-2}

=\frac{x-5+x-2}{x-2}

=\frac{2x-7}{x-2}

So, \frac{2x-7}{x-2}=\frac{3x}{x-1}

We now cross-multiply and expand:

(2x-7)(x-1)=3x(x-2)

2x^{2}-9x+7=3x^{2}-6x

Transposing: 3x^{2}-2x^{2}-6x+9x-7=0

x=\dfrac{-3\pm \sqrt{9-4\left( 1\right) \left( -7\right) }}{2\left( 1\right) }

x=\dfrac{-3\pm \sqrt{9+28}}{2}

x=\dfrac{-3\pm \sqrt{37}}{2}

Either x=\dfrac{-3 - \sqrt{37}}{2}

= -4.54

or

x=\dfrac{-3 + \sqrt{37}}{2}

= 1.54

to three significant figures.


Example – Full exam question

(a) Find the value of:

\dfrac{a+\sqrt{a^{2}+b^{2}}}{a^{2}-2ab}

When a=-4 and b=-3, give your answer as a fraction in its lowest term.

(b) Expand the brackets and simplify:

(3x^{2}-1)(2x+3)-x(9x-2)

(c)(i) Factorise 9x^{2}+5x-4

(ii) Use your answer to part (c)(i) to solve the equation 9x^{2}+5x-4=0

(d) The sum of three consecutive even numbers is 78. Find these three numbers.

(a) \dfrac{a+\sqrt{a^{2}+b^{2}}}{a^{2}-2ab}

When a=-4, b=-3, we obtain:

\dfrac{-4+\sqrt{\left( -4\right) ^{2}+\left( -3\right) ^{2}}}{}\\ \left( -4\right) ^{2}-2\left( -4\right) \left( -3\right)

= \dfrac{-4+\sqrt{16+9}}{16-24}

= \dfrac{-4+\sqrt{25}}{-8}

= \dfrac{-4+5}{-8}

= -\dfrac{1}{8}

(b) (3x^{2}-1)(2x+3)-x(9x-2)

= 6x^{3}+9x^{2}-2x-3-9x^{2}+2x

= 6x^{2}-3

(c)(i) 9x^{2}+5x-4

Find two numbers, sum = 5, product= 36.

1┃36

2┃18

3┃12

4┃9 → Good, choose 4,9.

9x^{2}+5x-4=9x^{2}-4x+9x-4

=x(9x-4)+1(9x-4)

= (x+1)(9x-4)

(ii) 9x^{2}+5x-4=0

(x+1)(9x-4)=0

Either x+1=0

= x = -1

or

9x-4=0

x=\frac{4}{9}

(d) Let the three even numbers be:

2x, 2x+2 and 2x+4, as they are consecutive.

Now, 2x+2x+2+2x+4=78

6x+6=78

6x=72

x=12

So that the three even numbers are:

2(12), 2(12)+2, 2(12)+4

= 24, 26, 28


Example – Exam question

The distance between London and York is 320 km. A train takes x hours to travel between London and York.

(a) Write down an expression, in terms of x, for the average speed of the train in km/h.

(b) A car takes 2 \frac{1}{2} hours longer than a train to travel between London and York.

The average speed of the train is 80 km/h greater than the average speed of the car.

Form an equation in x and show that it simplifies to 2x^{2}+5x-20=0.

(c) Solve the equation 2x^{2}+5x-20=0, giving your answer correct to two decimal places.

(d) Hence find the average speed of the car correct to the nearest km/h.

(a) London → York = 320 km

Average speed = \frac{Total distance}{Total time}

Average speed of the train = \frac{320 km}{x hr}

= \frac{320}{x} km/h

(b) Average speed of the car = \frac{320}{x+\frac{5}{2}} km/h

Average speed of the train average speed of the car = 80

\frac{320}{x}-\frac{320}{x+\frac{5}{2}} = 80

Divide by 80 throughout the equation:

\frac{4}{x}-\frac{4}{x+\frac{5}{2}}=1

Multiply by x(x+\frac{5}{2} on both sides

4(x+\frac{5}{2}) - 4x = 1 \times x(x+\frac{5}{2})

4x+10-4x = x(x+\frac{5}{2})

10=x^{2}+\frac{5}{2}x

Multiply by 2 on both sides

20 = 2x^{2}+5x

Rearranging gives 2x^{2}+5x-20=0

(c) Solving 2x^{2}+5x-20=0

x=\dfrac{-5\pm \sqrt{25-4\left( 2\right) \left( -20\right) }}{2\left( 2\right) }

=\dfrac{-5\pm \sqrt{185}}{4}

Either x=\dfrac{-5-\sqrt{185}}{4}

= -4.65 (two decimal places)

Or

x=\dfrac{-5+\sqrt{185}}{4}

= 2.15 (two decimal places)

(d) Average speed of the car = \frac{320}{x+\frac{5}{2}}

= \frac{320}{2.15037+2.5}

= \frac{320}{4.65037}

=68.8 km/h

69 km/h (nearest km/h)


Example

Solve the equation:

(2x+1)(3x-2)=49

We first need to rewrite the equation into the standard form ax^{2}+bx+c=0

(2x+1)(3x-2)=49

6x^{2}-4x+3x-2-49=0

6x^{2}-x-51=0

6x^{2}-18x+17x-51=0

6x(x-3)+17(x-3)=0

(6x+17)(x-3)=0

Either 6x+17=0

x=\frac{-17}{6}

Or

x-3=0

x=3